6.6 The Mean-Value Theorem for Derivatives

There is an important relation between the increment quotient $\frac{Δ y}{Δ x}$ and $\frac{d y}{d x} = f^{'} (x)$ . In calculus, many properties of functions can be deduced from this relation, which is called the Mean-Value Theorem. Let’s first consider the geometric significance of this theorem.

Consider the graph of $y = f (x)$ and two points $A (a, f (a))$ and $B (b, f (b))$ on it (Figure 1). Because $A G = b - a$ and $G B = f (b) - f (a)$ , the slope of the chord $A B$ is $\begin{matrix} (i) & \tan (∠ G A B) = \frac{G B}{A G} = \frac{f (b) - f (a)}{b - a} = \frac{Δ f}{Δ x} . \end{matrix}$

Figure 1: Geometric interpretation of the Mean-Value Theorem

As we can see, there is at least one point on the curve between $A$ and $B$ (as $P$ ) where the tangent to the curve $y = f (x)$ is parallel to the chord $A B$ . If the $x$ -value of $P$ is $c$ , the slope at $P$ is $\begin{matrix} (ii) & \tan θ = f^{'} (c) = \tan (∠ G A B) . \end{matrix}$ Equating (i) and (ii), we get $\begin{matrix} (iii) & \frac{f (b) - f (a)}{b - a} = f^{'} (c) . \end{matrix}$

In fact, we will find Eq. (iii) intuitive if we think of $f (t)$ as the position of a particle (that moves on a straight line) at time $t$ . The left-hand side of Eq. (iii) represents the average velocity in the time interval $[a, b]$ and the right-hand side of Eq. (iii) is the instantaneous velocity at some time. This equation states that at some instant the instantaneous velocity must be equal to the average velocity. For example, if the average speed (or average velocity) of a train is 100 km/h (kilometers per hour), then its speedometer must register 100 km/h at least once during the trip.

The formal statement of the mean-value theorem is as follows.

Theorem 1.The MEAN-VALUE THEOREM FOR DERIVATIVES. If function $f (x)$ is a function with the following properties:

$f$ is continuous on a closed interval $[a, b]$ and
$f$ is differentiable on the open interval $(a, b)$ ,

then there exists at least one point $c$ in the open interval $(a, b)$ such that $\begin{matrix} (iii) & f^{'} (c) = \frac{f (b) - f (a)}{b - a}, \end{matrix}$ or equivalently, $\begin{matrix} (iv) & f (b) - f (a) = f^{'} (c) (b - a) . \end{matrix}$

Show the proof …

The equation of the line that connects

A

and

B

y - f (a) = \frac{f (b) - f (a)}{b - a} (x - a)

y = f (a) + \frac{f (b) - f (a)}{b - a} (x - a) .

If we, as shown in Figure 2, subtract the right hand side of the above equation from

f (x)

and define a new function

h (x)

h (x) = f (x) - [f (a) + \frac{f (b) - f (a)}{b - a} (x - a)],

then

h (x)

satisfies the hypotheses of Rolle’s Theorem on

[a, b]

because

h

is (1) continuous on

[a, b]

, (2) differentiable on

(a, b)

, and (3)

h (a) = h (b) .

Therefore, there exists at least one point

c

in the interval

(a, b)

such that

h^{'} (c) = 0

. Because

h^{'} (x) = f (x) - \frac{f (b) - f (a)}{b - a},

we have

h^{'} (c) = f^{'} (c) - \frac{f (b) - f (a)}{b - a} = 0 \Rightarrow f^{'} (c) = \frac{f (b) - f (a)}{b - a} .

Figure 2: There might be more than one point that satisfies Eq. (c).

The Mean-Value Theorem for derivatives is sometimes called Lagrange’s Mean-Value Theorem.
Note that the Mean-Value Theorem does not assert where the mean value $c$ is located, except that it is somewhere between $a$ and $b$ . In fact, except for special cases, it is often difficult to determine this point. However, many important properties can be deduced from the mere existence of such a point.
Similar to Rolle’s theorem where $f^{'}$ vanishes, there might be more than one point that satisfies the Mean-Value Theorem. For example, see Figure 3 where two points satisfy Eq. (iii).
If at any point in the interval $(a, b)$ , the derivative $f^{'}$ fails to exists, the Mean-Value Theorem may not hold true.

Figure 3: There might be more than one point that satisfies Eq. (c).

We can express Eq. (iii) somewhat differently by noticing that the number $c$ can be written as $c = a + θ (b - a),$ where $θ$ is a certain number between 0 and 1 ( $0 < θ < 1$ ). Now if we replace $a$ by $x$ and $b$ by $x + h$ , we can express the Mean-Value Theorem by the formula $\frac{f (x + h) - f (x)}{h} = f^{'} (x + θ h),$ or $f (x + h) = f (x) + h f^{'} (x + θ h) where 0 < θ < 1.$

Example 1

Verify the conclusion of the Mean-Value Theorem for

f (x) = x^{2} - 4 x + 3

on the interval

[- 2, 3]

Solution 1

Because

f

is a polynomial, it is continuous and differentiable for all values of

x

. Therefore, the hypotheses of the Mean-Value Theorem are satisfied for any closed interval. The slope of the secant line through the endpoint values (or the average change of

f

over the interval

[- 2, 3]

) is

$\frac{f (3) - f (- 2)}{3 - (- 2)} = \frac{0 - 15}{5} = - 3.$ The slope of the tangent line (or the instantaneous change of $f$ ) is $f^{'} (x) = 2 x - 4$ The Mean-Value Theorem says there exists some $c$ in $(- 2, 3)$ such that $f^{'} (c) = - 3$ : $f^{'} (c) = 2 c - 4 = - 3$ $\Rightarrow c = \frac{1}{2} .$ Because $1 / 2$ lies in the interval $(- 2, 3)$ , the $c$ value referred to in the conclusion of the Mean-Value Theorem is $c = 1 / 2$ . See Figure 4.

Figure 4

Example 2

Show that
(a) If

- 1 \leq f^{'} (x) \leq 1

for every

x

, then

| f (x) - f (y) | \leq | x - y |

for every

x

and

y

.
(b) For every

x

and

y

| \sin x - \sin y | \leq | x - y | .

Solution 2

(a) The existence of

f^{'} (x)

shows that

f

is differentiable and consequently continuous everywhere (see here). Therefore, the hypotheses of the Mean-Value Theorem are satisfied, and there exists some

c

between

x

and

y

such that

f (x) - f (y) = f^{'} (c) (x - y) .

Taking the absolute value of both sides:

| f (x) - f (y) | = | f^{'} (c) | | x - y | .

Because

- 1 \leq f^{'} (x) \leq 1

for every

x

| f^{'} (c) | \leq 1

and thus

| f (x) - f (y) | \leq | x - y | .

(b) Let

f (x) = \sin x

. Then

- 1 \leq f^{'} (x) = \cos x \leq 1.

It follows from part (a) that

| f (x) - f (y) | \leq | x - y |

$| \sin x - \sin y | \leq | x - y | .$

Example 3

Show that if

x > 0

\sqrt{1 + x} < 1 + \frac{1}{2} x

Solution 3

Let

f (x) = \sqrt{1 + x}

. Then

f^{'} (x) = \frac{1}{2 \sqrt{1 + x}} .

The function

f

is continuous and differentiable for all

x > - 1

. Let

a = 0

and

b = x > 0

. By the Mean-Value Theorem, there is a number

c

between

0

and

x

such that

f (x) - f (0) = f^{'} (c) (x - 0)

\begin{matrix} (i) & \sqrt{1 + x} - 1 = f^{'} (c) x \end{matrix}

Now we claim that

f^{'} (c) < \frac{1}{2}

, because for

c > 0

1 < \sqrt{1 + c}

and hence

\frac{1}{\sqrt{1 + c}} < 1

\begin{matrix} (ii) & f^{'} (c) = \frac{1}{2 \sqrt{1 + c}} < \frac{1}{2} . \end{matrix}

Combining (i) and (ii), we obtain

\sqrt{1 + x} - 1 < \frac{1}{2} x

\sqrt{1 + x} < 1 + \frac{1}{2} x .

Theorem 2. CAUCHY’S MEAN-VALUE FORMULA. Assume $f (x)$ and $g (x)$ are continuous on $[a, b]$ and are differentiable on $(a, b)$ . Then there exists at least one point $c$ in $(a, b)$ such that $f^{'} (c) [g (b) - g (a)] = g^{'} (c) [f (b) - f (a)] .$ If $g^{'} (c) \neq 0$ and $g (a) \neq g (b)$ , this is equivalent to: $\frac{f (b) - f (a)}{g (b) - g (a)} = \frac{f^{'} (c)}{g^{'} (c)} .$

Show the proof …

Similar to the proof of the Mean-Value theorem, we define a new function:

h (x) = f (x) [g (b) - g (a)] - g (x) [f (b) - f (a)] .

Because

f (x)

and

g (x)

are both continuous in

[a, b]

and differentiable in

(a, b)

, and

g (b) - g (a)

and

f (b) - f (a)

are just two numbers, it follows from Theorem 1 in Section on Continuity and Theorem 2 in Section on Differentiation Rules that

h (x)

is also continuous on

[a, b]

and differentiatiable on

(a, b)

. We note that

h (a) = h (b) = f (a) [g (b) - g (a)] - g (a) [f (b) - f (a)] = f (a) g (b) - f (b) g (a) .

Therefore, the conditions of Rolle’s theorem are satisfied. Applying Rolle’s theorem to

h (x)

, we find that for some

c

(a, b)

, we have

h^{'} (c) = f^{'} (c) [g (b) - g (a)] - g^{'} (c) [f (b) - f (a)] = 0,

f^{'} (c) [g (b) - g (a)] = g^{'} (c) [f (b) - f (a)] .

Notice that the Mean-Value Theorem (Lagrange’s theorem) is the special case of Cauchy’s Mean-Value Formula when $g (x) = x$ .

Show the proof …

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Show the proof …

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